91Ó°ÊÓ

$E$and $F$independent

$E$and $G$independent

$⇒E$independent of$F∪G$

This is false

Counterexample: Two fair dice are rolled.

$E$- the sum of the results is even.

$F$- result on the second die is $3.$

$G$- result on the first die is$4$

Since $P(Eâˆ\pounds F)=\frac{1}{2}$, and $P(Eâˆ\pounds G)=\frac{1}{2}$and $P\left(E\right)=\frac{1}{2}$(this can be obtained by conditioning on the number on the first die).

E and F and E and G are independent.

$P(Eâˆ\pounds F∪G)=\frac{6}{11}$- method of counting.

Thus:

$P(Eâˆ\pounds F∪G)≠P\left(E\right)$

.So $E$and $F∪G$are not independent.

$E$and $F$independent

$E$and $G$independent

$Fâˆ\copyright G=\mathrm{âˆ\dots }$

$⇒E$independent of $F∪G$

A and B are independent $⇔P(Aâˆ\pounds B)=P\left(A\right)/P(Bâˆ\pounds A)=P\left(B\right)$

$P(Eâˆ\pounds F∪G)=\frac{P(Eâˆ\copyright (F∪G\left)\right)}{P(F∪G)}$

$=\frac{P(EF∪EG))}{P(F∪G)}$

$Fâˆ\copyright G=\mathrm{âˆ\dots },⇒EFâˆ\copyright EG=\mathrm{âˆ\dots }$

$=\frac{P\left(EF\right)+P\left(EG\right)}{P\left(F\right)+P\left(G\right)}$

Independence

$=\frac{P\left(E\right)P\left(F\right)+P\left(E\right)P\left(G\right)}{P\left(F\right)+P\left(G\right)}$

$=\frac{P\left(E\right)\left[P\right(F)+P(G\left)\right]}{P\left(F\right)+P\left(G\right)}$

$=P\left(E\right)$

And$P(Eâˆ\pounds F∪G)=P\left(E\right)$proves independence

E and F independent

F and G independent

E and FG independent

$Fâˆ\copyright G=\mathrm{âˆ\dots }$

This is correct

Use

A and B are independent $⇔P\left(AB\right)=P\left(A\right)P\left(B\right)$

$P\left(EFG\right)=P\left(E\right)P\left(FG\right)$

E,FG independent

$=P\left(E\right)P\left(F\right)P\left(G\right)$

F,G independent

$=P\left(EF\right)P\left(G\right)$

E,F independent

And $P(EFG)=P\left(G\right)P(EF)$proves independence of G and E F